Balancing Redox Equations

It is not always possible to balance redox
equations using the simple inspection technique. The following unbalanced net ionic equation provides an example.

Au3+(aq)
+ I-(aq)
-->Au(s)
+ I2(s)

At first glance, it seems that this
equation can be balanced by placing a 2 in front of the I-.

Au3+(aq)
+ 2I-(aq)
-->Au(s) + I2(s)

Note, however, that although the atoms are
now balanced, the charge is not. The sum of the charges on the left is +1,
and the sum of the charges on the right is zero, as if the products could
somehow have one more electron than the reactants. To correctly balance
this equation, it helps to look more closely at the oxidation and
reduction that occur in the reaction. The iodine atoms are changing their
oxidation number from -1 to 0, so each iodide ion must be losing one
electron. The Au3+ is changing to Au, so each gold(III) cation
must be gaining three electrons. The half-reactions are:

I-(aq)
-->I2(s) + e-

Au3+(aq) + 3e-
-->Au(s)

We know that in redox reactions, the number
of electrons lost by the reducing agent must be equal to the number of
electrons gained by the oxidizing agent; thus, for each Au3+
that gains three electrons, there must be three I- ions that
each lose one electron. If we place a 3 in front of the I- and
balance the iodine atoms with a 3/2 in front of the I2, both
the atoms and the charge will be balanced.

Au3+(aq) + 3I-(aq)
-->Au(s)
+ 3/2I2(s)

or 2Au3+(aq)
+6I-(aq)
-->2Au(s) + 3I2(s)

Click on the links below to read about the three general
techniques for balancing redox equations.